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Introduction of VSEPR

N2 has greate...

Question

$N_{2}$ has greater dissociation energy than ${N_{2}}^{+}$ , where as $O_{2}$ has a lower dissociation energy than ${O_{2}}^{+}$ because:

bond order is reduced when

O_{2}

is ionized to

{O_{2}}^{+}

and bond order is increased when

N_{2}

is ionized to

{N_{2}}^{+}

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bond order is increased when

O_{2}

is ionized to

{O_{2}}^{+}

and bond is decreased when

N_{2}

is ionized to

{N_{2}}^{+}

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bond order is decreased when

O_{2}

is ionized to

{O_{2}}^{+}

and bond order is decreased when

N_{2}

is ionized to

{N_{2}}^{+}

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none of the above

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Solution

The correct option is B bond order is increased when $O_{2}$ is ionized to ${O_{2}}^{+}$ and bond is decreased when $N_{2}$ is ionized to ${N_{2}}^{+}$ .

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Similar questions

Q. When

N_{2}^{+}

is form

N_{2}

bond-order ......... and when

O_{2}^{+}

is formed from

O_{2}

bond-order ........

Q. Explain why

N_{2}

has greater bond dissociation energy than

N_{2}^{+}

whereas

O_{2}

has lesser bond dissociation energy than

O_{2}^{+}

N_{2}

has greater bond energy than

{N_{2}}^{+}

but

O_{2}

has tower bond dissociation energy then

{O_{2}}^{+}

. Explain.

Q. When

N_{2}

goes to

N_{2}^{+}

, the N-N bond distance ..... and when

O_{2}

goes to

O_{2}^{+}

, the O-O bond distance ....

When $N_{2}$ goes to $N_{2}^{+}$ , the N - N bond distance ... and

when $O_{2}$ goes to $O_{2}^{+}$ , the o - o bond distance....

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N2 has greater dissociation energy than N2+, where as O2 has a lower dissociation energy than O2+ because:

The correct option is B bond order is increased when O2 is ionized to O2+ and bond is decreased when N2 is ionized to N2+.

$N_{2}$ has greater dissociation energy than ${N_{2}}^{+}$ , where as $O_{2}$ has a lower dissociation energy than ${O_{2}}^{+}$ because:

The correct option is B bond order is increased when $O_{2}$ is ionized to ${O_{2}}^{+}$ and bond is decreased when $N_{2}$ is ionized to ${N_{2}}^{+}$ .